Problem 37 - Intermediate Algebra

Question

Solve each equation. Give the exact solution and an approximation to four decimal places. $$ e^{2.9 x}=4.5 $$

Step-by-Step Solution

Answer

The exact solution is \(x = \frac{\ln(4.5)}{2.9}\) and the approximation is \(x \approx 0.5186\).

Step 1: Take the Natural Logarithm

To isolate the variable \(x\), take the natural logarithm (ln) of both sides of the equation. \[\ln(e^{2.9x}) = \ln(4.5)\] Using the property of logarithms that states \(\ln(e^y) = y\), this simplifies to: \[2.9x = \ln(4.5)\]

Step 2: Solve for x

Now that the equation is simplified to \(2.9x = \ln(4.5)\), divide both sides by 2.9 to solve for \(x\): \[x = \frac{\ln(4.5)}{2.9}\]

Step 3: Calculate the Exact Solution

Calculate \(\ln(4.5)\) using a calculator. \(\ln(4.5) \approx 1.5041\). Now substitute back into the expression for \(x\):\[x = \frac{1.5041}{2.9} \] This simplifies to: \[x \approx 0.5186\]

Step 4: Present the Solution

The exact solution for \(x\) is \(\frac{\ln(4.5)}{2.9}\). The approximation of this solution, rounded to four decimal places, is \(x \approx 0.5186\).

Key Concepts

Natural Logarithms

Natural logarithms are used extensively in solving exponential equations. The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. This base is preferred in calculus and many areas of algebra because it simplifies the derivatives and integrals of exponential functions.
\[ \ln(e^y) = y \] is a key property used when solving equations like \( e^{2.9x} = 4.5 \). Taking the natural logarithm of both sides helps to expose the exponent, making it possible to isolate the variable \( x \) and solve the equation. Let us simplify this scenario by breaking down the steps:

• Taking \( \ln \) of \( e^{2.9x} \) simplifies to \( 2.9x \)
• Taking \( \ln(4.5) \) involves using calculators or log tables, giving an approximate value

Approximation

When solving exponential equations, especially those involving logarithms, we often round the solution for practical purposes. In mathematical calculations, you'll frequently see approximations to a certain number of decimal places. In our current scenario, we need to approximate the solution to four decimal places.
Approximation is helpful for:

• Providing a quick and easy-to-understand result without requiring an exact number which might be complex
• Applying mathematical results in real-world scenarios where absolute precision is unnecessary or impractical

In the example provided, after calculating \( \ln(4.5) \) as 1.5041, dividing by 2.9 gives us an approximate value of \( x \approx 0.5186 \), which we round to four decimal places for simplicity and clarity.

Exact Solution

The exact solution to an exponential equation is an expression that can be interpreted without rounding. This keeps all decimal points intact and is expressed symbolically when exact numerical computation methods are not feasible. In the case of the equation \( e^{2.9x} = 4.5 \), to find the exact solution, we rewrite it as:
\[ x = \frac{\ln(4.5)}{2.9} \]Using this formulation allows for perfect adaptability when subjected to further algebraic manipulations where precision is paramount. This is especially useful in theoretical derivations or computer calculations needing full accuracy.

Algebra

Solving exponential equations often involves essential algebraic techniques to isolate and solve for unknown variables. Algebra provides the toolkit needed to rearrange equations and resolve various forms of mathematical expressions. In the given example, algebra is used to:

• Take the natural logarithm of both sides to simplify the exponential form into a linear expression
• Divide both sides by the coefficient of \( x \) to isolate \( x \)

This process involves understanding operations such as multiplication, division, and applying specific logarithmic properties. Reiterating these basic but essential algebraic steps ensures mastery when dealing with more complex equations or scenarios.